Complete the following statement. Round your answer to the nearest cent.<br><br> 5% of $585.26 = $

Ms. Carson's art class made a large rectangular mural. The length of the mural is 9 feet, and it's area is 72 square feet. What

lorasvet [3.4K]

You do the reverse and divide the 2 values to find out the last one

72/9 = 8

Do you get it?

8 0

3 years ago

Read 2 more answers

The students in Mr. Wilson's Physics class are making golf ball catapults. The

Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is $0 \leq x \leq 48.6\ ft$ and the range is $0 \leq y \leq 8.3\ ft$

Step-by-step explanation:

we have

$y=-0.014x^{2} +0.68x$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's distance from the catapult in feet

y is the flight of the balls in feet

Part a) How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-0.014x^{2} +0.68x=0$

so

$a=-0.014\\b=0.68\\c=0$

substitute in the formula

$x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}$

$x=\frac{-0.68(+/-)0.68} {(-0.028)}$

$x=\frac{-0.68(+)0.68} {(-0.028)}=0$

$x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft$

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

$y=-0.014x^{2} +0.68x$

Find out the derivative and equate to zero

$0=-0.028x +0.68$

Solve for x

$0.028x=0.68$

$x=24.3$

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint between the x-intercepts

$x=(0+48.6)/2=24.3\ ft$

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

$y=-0.014(24.3)^{2} +0.68(24.3)$

$y=8.3\ ft$

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

Part c) What is a reasonable domain and range for this function?

we know that

A reasonable domain is the distance between the two x-intercepts

so

$0 \leq x \leq 48.6\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

so

we have the interval -----> [0,8.3]

$0 \leq y \leq 8.3\ ft$

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

8 0

3 years ago

How do you increase £480 by 7.5%

elena55 [62]

$\sf 480+ \frac{7,5*48\not0}{10\not0}= 480+ \frac{36\not0}{1\not0}=480+36=\boxed{\sf 516~\£}$

4 0

3 years ago

Please what is 33×33+33-33÷0

DochEvi [55]

Answer:

33×33+33-33÷0=infinity

33×33=1089

33-33=0

Taking L.H.S

=33×33+(33-33)÷0

=1089+0÷0

=1089÷0

=infinity

L.H.S=R.H.S

Step-by-step explanation:

4 0

3 years ago

Seirra found that 15 meters is equivalent to 49.2 feet which of the following is an equivalent ratio

loris [4]

A.........................................

8 0

3 years ago

Read 2 more answers

Complete the following statement. Round your answer to the nearest cent. 5% of $585.26 = $